String Matrix to Character Matrix
String Matrix to Character Matrix
PROBLEM STATEMENT :
The program must accept a matrix of size RxC containing only words as the input. All words in the matrix have the same length as 9. The program must form a character matrix of size (R*3)x(C*3) by expanding the each word to a 3*3 character matrix based on the following conditions. - The 1st row of the 3*3 matrix must be formed using the first 3 characters of the word. - The 2nd row of the 3*3 matrix must be formed using the middle 3 characters of the word. - The 3rd row of the 3*3 matrix must be formed using the last 3 characters of the word. Finally, the program must print the character matrix as the output.
Boundary Condition(s):
1 <= R, C <= 50
Input Format:
The first line contains R and C separated by a space.
The next R lines, each contains C words separated by a space.
Output Format:
The first R*3 lines, each contains C*3 characters separated by a space.
Example Input/Output 1:
Input: ()
3 3
skillrack chocolate IMPORTANT
Knowledge Seventeen vegetable
Crocodile nutrition Fireboard
Output:
s k i c h o I M P
l l r c o l O R T
a c k a t e A N T
K n o S e v v e g
w l e e n t e t a
d g e e e n b l e
C r o n u t F i r
c o d r i t e b o
i l e i o n a r d Explanation:
Here R = 3 and C = 3.
The word skillrack is expanded to a 3x3 character matrix as given below.
l l r
a c k
The word chocolate is expanded to a 3x3 character matrix as given below.
c o l
a t e
Similarly the remaining words are expanded to 3x3 character matrices.
Hence the output is
l l r c o l O R T
a c k a t e A N T
K n o S e v v e g
w l e e n t e t a
d g e e e n b l e
C r o n u t F i r
c o d r i t e b o
i l e i o n a r d
Example Input/Output 2:
Input: ()
3 4
Challenge irregular PINEAPPLE halloween
affection NUTRITION Crocodile beginning
CONFUSION celebrity Delicious LIGHTNING
Output:
C h a i r r P I N h a l
l l e e g u E A P l o w
n g e l a r P L E e e n
a f f N U T C r o b e g
e c t R I T c o d i n n
i o n I O N i l e i n g
C O N c e l D e l L I G
F U S e b r i c i H T N
I O N i t y o u s I N G
SOLUTION :
C (Programming Language)
#include<stdio.h>
#include<stdlib.h>
int main()
{
int i,j;
scanf("%d %d\n",&i,&j);
char a[i][j][9];
for(int p=0;p<i;p++,scanf("\n"))
{
for(int r=0;r<j;r++)
{
scanf("%s ",a[p][r]);
}
}
for(int p=0;p<i;p++)
{
for(int s=0;s<9;s+=3)
{
for(int r=0;r<j;r++)
{
for(int c=s;c<s+3;c++)
{
printf("%c ",a[p][r][c]);
}
}
printf("\n");
}
}
}
C++ (CPP)
#include <bits/stdc++.h>
using namespace std;
int main(int argc, char** argv)
{
int i,j;
cin>>i>>j;
char a[i][j][9];
for(int p=0;p<i;p++,scanf("\n"))
{
for(int r=0;r<j;r++)
{
cin>>a[p][r];
}
}
for(int p=0;p<i;p++)
{
for(int s=0;s<9;s+=3)
{
for(int r=0;r<j;r++)
{
for(int c=s;c<s+3;c++)
{
cout<<a[p][r][c]<<" ";
}
}
cout<<"\n";
}
}
}
JAVA
import java.util.*;
public class Hello {
public static void main(String[] args) {
Scanner sc=new Scanner(System.in);
int R=sc.nextInt(),C=sc.nextInt();
String[][] matrix=new String[R][C];
for(int row=0;row<R;row++){
for(int col=0;col<C;col++){
matrix[row][col]=sc.next();
}
}
for(int row=0;row<R;row++){
for(int ind=0;ind<9;ind+=3){
for(int col=0;col<C;col++){
for(int currInd=ind;currInd<ind+3;currInd++){
System.out.print(matrix[row][col].charAt(currInd)+" ");
}
}
System.out.print("\n");
}
}
}
}
PYTHON
r,c=map(int,input().split())
m=[input().split() for _ in range(r)]
for i in range(r):
for k in range(0,9,3):
for j in range(c):
print(*m[i][j][k:k+3],end=' ')
print()
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